Question

A party rental company has chairs and tables to rent.There were two customers who rented both chairs and tables last week.The table below shows the number of chairs,the number of tables,and the total cost (in dollars) for those two customers.Let x be the cost to rent a chair.Let y be the cost to rent a table.How much does each chair and each table cost to rent?

Answer 1

We can set up two equations with the given information.

Let x be the cost to rent a chair.

Let y be the cost to rent a table.

$\begin{gathered} 5x+3y=34\quad eq.1 \\ 7x+9y=80\quad eq.2 \end{gathered}$

Now we can solve these two equations using the substitution method.

First, separate out one variable from either of the two equations.

$\begin{gathered} 5x+3y=34 \\ 5x=34-3y \\ x=\frac{34-3y}{5} \end{gathered}$

Now substitute it into the other equation

$\begin{gathered} 7x+9y=80 \\ 7(\frac{34-3y}{5})+9y=80 \\ \frac{238-21y}{5}+9y=80 \\ \frac{238-21y+5(9y)}{5}=80 \\ \frac{238-21y+45y}{5}=80 \\ 238-21y+45y=5\cdot80 \\ 238+24y=400 \\ 24y=400-238 \\ 24y=162 \\ y=\frac{162}{24} \\ y=6.75 \end{gathered}$

So, we got the cost of a table that is y = $6.75

Now substitute it into the above equation to find x.

$\begin{gathered} x=\frac{34-3(6.75)}{5} \\ x=\frac{34-20.25}{5} \\ x=\frac{13.75}{5} \\ x=2.75 \end{gathered}$

So, we got the cost of a chair that is x = $2.75

The cost to rent a chair = x = $2.75

The cost to rent a table = y = $6.75